Lecture 0503 Calorimetry and Hess s Law

Transcription

1 Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Lecture 0503 Calorimetry and Hess s Law John D. Bookstaver St. Charles Community College Cottleville, MO 2009, Prentice-Hall, Inc.

2 Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

3 Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1 C) is its heat capacity.

4 Heat Capacity and Specific Heat We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

5 Heat Capacity and Specific Heat Specific heat, then, is heat transferred Specific heat = mass temperature change c p = q m T

6 Problem a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g- K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 C. (b) What temperature change would these rocks undergo if they emitted 450 kj of heat?

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8 Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

9 Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation: q = m c p T

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11 Bomb Calorimetry Reactions can be carried out in a sealed bomb such as this one. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction.

12 Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. For most reactions, the difference is very small.

13 Problem A g sample of lactic acid (HC 3 H 5 O 3 ) is burned in a calorimeter whose heat capacity is kj/ C. The temperature increases from C to C. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole.

14 Hess s Law H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested. However, we can estimate H using published H values and the properties of enthalpy.

15 Hess s Law Hess s law states that [i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.

16 Hess s Law Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

17 Problem Calculate ΔH for the reaction NO(g) + O(g) NO 2 (g) given the following information:

18 Enthalpies of Formation An enthalpy of formation, H f, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

19 Standard Enthalpies of Formation Standard enthalpies of formation, H f, are measured under standard conditions (25 C and 1.00 atm pressure).

20 Calculation of H C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) Imagine this as occurring in three steps: C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l)

21 Calculation of H C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) Imagine this as occurring in three steps: C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l)

22 Calculation of H C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) Imagine this as occurring in three steps: C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l)

23 C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) The sum of these equations is: Calculation of H C 3 H 8 (g) 3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g) 3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g) 4 H 2 O (l) C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l)

24 Calculation of H We can use Hess s law in this way: H = n H f products m H f reactants where n and m are the stoichiometric coefficients.

25 Calculation of H C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) H = [3( kj) + 4( kj)] [1( kj) + 5(0 kj)] = [( kj) + ( kj)] [( kj) + (0 kj)] = ( kj) ( kj) = kj

26 Problem Given the following standard enthalpy change, use the standard enthalpies of formation in Table 5.3 to calculate the standard enthalpy of formation of CuO(s):

27 Problem 5.7 (cont.) (a) Write a balanced chemical equation for the decomposition of trinitroglycerin. (b) Calculate the standard heat of formation of trinitroglycerin.

28 Problem 5.7 (cont.) (c) A standard dose of trinitroglycerin for relief of angina is 0.60 mg. If the sample is eventually oxidized in the body (not explosively, though!) to nitrogen gas, carbon dioxide gas, and liquid water, what number of calories is released? (d) One common form of trinitroglycerin melts at about 3 C. From this information and the formula for the substance, would you expect it to be a molecular or ionic compound? Explain. e) Describe the various conversions of forms of energy when trinitroglycerin is used as an explosive to break rockfaces in highway construction.

29 Types of Heats

30 Types of Heats

31 Types of Heats

32 Types of Heats

33 Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats.

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35 Energy in Fuels The vast majority of the energy consumed in this country comes from fossil fuels.

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37 Energy in Fuels